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The probability of a four-of-a-kind when choosing five cards from a deck, example 58.5
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- Опубликовано: 9 окт 2019
- In this video, we solve a problem that seems simple at first glance, but for many students the problem is a challenge. What is the probability you draw a four-of-a-kind when choosing five cards from a deck of 52 cards (for example: 4 kings and the 3 of hearts is a four-of-a-kind hand)?
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I was searching like four days in row about a good explanation.. and here we are.. thank you so much professor
You’re welcome, and thanks for the kind words!
Thank you for the explanation. Spent and embarrassing amount of time looking at this problem with no idea what to do.
It’s not an easy problem, so it’s normal to find it a challenge.
i haven't finished watching this lecture to the end, I already gave it a like. beautiful explanation!
Wonderful!
Very well explained. thinking through the problem is excellent. The explanation of the last choice of 48 possibilities in both ways is great.
Thank you!
Thank you, I learned a lot. I was recently trying to get into probability math and this really helped me. Thank you!
Great to hear!
Dude! You're a freaking hero.
Happy I could help!
WOW! the way that you explain it's awesome!!! thanks a lot. Saludos desde Mexico!
Thank you 😊 I’m glad it was helpful. I have close friends from Cuernavaca 🇲🇽
Thank you.
Super clear explanation! thank you so much for the video...
You’re welcome
If I have three of a kind... What probability is best.. to get rid of one card or two cards?
👍
thanks man
You’re welcome
How can I calculate the probability of getting 4 kings when choosing 13 cards?
Assuming order doesn’t matter, it should be (4C4)*(48C9)/(52C13)
In case your calc freezes up, this should reduce to 13*12*11*10/(52*51*50*49) double check this though bc right now I don’t have a proper pencil and paper
@@dmcguckian How about the probability of getting 2 aces when picking 13 cards from the deck? I don't really understand how you got (4C4)*(48C9).
@@dmcguckian On your other video titled "The Probability of a Four-of-a-Kind", you used 13 and 48
@@Megaheropap think about the fact that the 4 kings is a special case of the four of a kind (if we were choosing 5 cards). There would be 13 of these hands. That’s how you get from four kings to a generic four of a kind
I'm having a hard time finding out the probability for the following:
suited cards: A-K-Q-J
This would be out of 7 cards to play, from a normal 52 card deck.
As well as any 4 card straight flush with 7 cards, from a 52 card deck.
Would it be:
(4|1)(52|4)(48|1)(47|1)(46|1) / 52C7
?
Just off the top of my head, I would say there are 4 ways to get A,K,Q,J of the same suite. Then you need to pick the remaining three cards carefully because if you want the other cards to not create a hand, you need to avoid selecting any 10s. You need to avoid any cards of the same suite. You need to avoid any pairs, so no A, K, Q, J, and no match among the cards you select that are not 10s, As, Ks, Qs, Js. Once you carefully construct the list of acceptable options the rest should follow easily.
Thankyou very much
Does this apply to poker? Because I got 4 cards with the same picture.
It's supposed to be for of a kind Wright?
Yes, this is the calculation for the probability of randomly selecting 5 cards from a shuffled deck and choosing a four of a kind. This is not that same as the probability of getting four of a kind in the normal play of a game of poker because you get to swap out cards even in a single hand of poker, let alone in a full game with several hands.
Confusion sr
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